TITLE: The kernel of all invariant means
QUESTION [7 upvotes]: Let $G$ be a discrete group which is amenable (i.e. it admits an left-invariant mean, i.e. a continuous positive normalised linear functional $m:\ell^\infty(G) \to \mathbb{R}$ such that $\forall g \in G, m \circ \lambda_g = m$ where $\lambda_g : \ell^\infty(G) \to \ell^\infty(G)$ is the left-regular action of $G$). Let $\mathcal{M}$ be the set of all invariant means on $G$.
Let $c_0(G)$ be the space of bounded functions decreasing to $0$ at $\infty$ (i.e. $f \in c_0(G)$ if $f$ is in the closure of the space of finitely supported functions, i.e. $f \in \ell^\infty(G)$ and $\forall \epsilon >0$ there exists a finite set $F \subset G$ with $\lVert f\rVert_{\ell^\infty(G \setminus F)} \leq \epsilon$).
It's not difficult to see that $c_0(G)$ is included in the kernel of all (left- or right- or bi-)invariant mean, provided $G$ is infinite. When $G$ is finite, there is only one such mean, and it is trivial to compute the kernel. However the cardinality of $\mathcal{M}$ is otherwise quite big (if I remember correctly, it is uncountable for $G=\mathbb{Z}$).
Question: Assuming $G$ is infinite, is there a description/characterisation of the elements of $\ell^\infty(G)$ which belong to the kernel of all the invariant means, i.e. $\bigcap_{m \in \mathcal{M}} \ker m$?

REPLY [2 votes]: (Essentially from Narutaka Ozawa's comments)
Let $K = \bigcap_{m \in \mathcal{M}} \ker m$. The trivial answer would be that $K$ is the closed linear span of
$$S =\{f-\lambda_g(f):\;g\in G,\;f\in\ell^\infty(G) \}.$$
For a non-amenable group this is all of $\ell^\infty(G)$. For an amenable group with Følner net $(F_i)$, then for any $f \in K$
$$\lim_{i\to\infty}\sup_{x\in G}\frac{\left|\sum_{y\in F_ix}f(y)\right|}{|F_i|}=0.$$
Indeed, if it's nonzero, we can find $(x_i)$ such that
$$\limsup\frac{\left|\sum_{y\in F_ix_i}f(y)\right|}{|F_i|}>0;$$
then the uniform measure on $F_ix_i$ accumulates to an invariant mean that is non-zero at $f$.
Hence, if $f \in K$, then
$$f_i(x)=\sum_{y\in F_ix}\frac{f(y)}{|F_i|}$$
has the same mean as $f$ for every invariant mean. A consequence of the above is that $\|f_i\|_{\ell^\infty} \to 0$. Thus $f - f_i \in S$ and $f-f_i \to f$ which shows that $K \subset \overline{S}$.
The inclusion $\overline{S} \subset K$ is a direct consequence of the invariance of the mean.